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Metamath Proof Explorer

Theorem reixi

Description: ixi without ax-mulcom . (Contributed by SN, 5-May-2024)

Ref Expression
Assertion reixi ( i · i ) = ( 0 −ℝ 1 )

Proof

Step Hyp Ref Expression
1 ax-i2m1 ⊢ ( ( i · i ) + 1 ) = 0
2 1re ⊢ 1 ∈ ℝ
3 renegid2 ⊢ ( 1 ∈ ℝ → ( ( 0 −ℝ 1 ) + 1 ) = 0 )
4 2 3 ax-mp ⊢ ( ( 0 −ℝ 1 ) + 1 ) = 0
5 1 4 eqtr4i ⊢ ( ( i · i ) + 1 ) = ( ( 0 −ℝ 1 ) + 1 )
6 ax-icn ⊢ i ∈ ℂ
7 6 6 mulcli ⊢ ( i · i ) ∈ ℂ
8 7 a1i ⊢ ( ⊤ → ( i · i ) ∈ ℂ )
9 rernegcl ⊢ ( 1 ∈ ℝ → ( 0 −ℝ 1 ) ∈ ℝ )
10 9 recnd ⊢ ( 1 ∈ ℝ → ( 0 −ℝ 1 ) ∈ ℂ )
11 2 10 mp1i ⊢ ( ⊤ → ( 0 −ℝ 1 ) ∈ ℂ )
12 1cnd ⊢ ( ⊤ → 1 ∈ ℂ )
13 8 11 12 sn-addcan2d ⊢ ( ⊤ → ( ( ( i · i ) + 1 ) = ( ( 0 −ℝ 1 ) + 1 ) ↔ ( i · i ) = ( 0 −ℝ 1 ) ) )
14 5 13 mpbii ⊢ ( ⊤ → ( i · i ) = ( 0 −ℝ 1 ) )
15 14 mptru ⊢ ( i · i ) = ( 0 −ℝ 1 )